Question: The point $O$ is the center of the circle circumscribed about $\triangle ABC$, with $\angle BOC = 120^{\circ}$ and $\angle AOB =
140^{\circ}$, as shown. What is the degree measure of $\angle
ABC$?

[asy]
pair A,B,C;
draw(Circle((0,0),20),linewidth(0.7));
label("$O$",(0,0),S);
A=(-16,-12);
C=(16,-12);
B=(3,19.7);
draw(A--B--C--cycle,linewidth(0.7));
label("$140^{\circ}$",(0,0),W);
label("$120^{\circ}$",(0,0.3),E);
draw(C--(0,0)--B);
draw(A--(0,0));
label("$A$",A,SW);
label("$B$",B,NE);
label("$C$",C,SE);
[/asy]
Explanation: Since $OA=OB=OC$, triangles $AOB$, $BOC$, and $COA$ are all isosceles. Hence \[
\angle ABC = \angle ABO + \angle OBC =
\frac{180^{\circ}-140^{\circ}}{2}+
\frac{180^{\circ}-120^{\circ}}{2}=\boxed{50^{\circ}}.
\]OR


Since \[
\angle AOC = 360^{\circ}-140^{\circ}-120^{\circ}=100^{\circ},
\]the Central Angle Theorem implies that \[
\angle ABC = \frac{1}{2}\angle AOC = \boxed{50^{\circ}}.
\]